648 CHAPTER 32. MEASURES AND INTEGRALS

32.10 Integrals of Real Valued FunctionsAs long as you are allowing functions to take the value +∞, you cannot consider somethinglike f +(−g) and so you can’t very well expect a satisfactory statement about the integralbeing linear until you restrict yourself to functions which have values in a vector space. Tobe linear, a function must be defined on a vector space. The integral of real valued functionsis next.

Definition 32.10.1 Let (Ω,F ,µ) be a measure space and let f : Ω → R be mea-surable. Then it is said to be in L1 (Ω,µ) when∫

Ω

| f (ω)|dµ < ∞

Lemma 32.10.2 If g−h = ĝ− ĥ where g, ĝ,h, ĥ are measurable and nonnegative, withall integrals finite, then ∫

Ω

gdµ −∫

Ω

hdµ =∫

Ω

ĝdµ −∫

Ω

ĥdµ

Proof: From Theorem 32.9.1,∫ĝdµ +

∫hdµ =

∫(ĝ+h)dµ =

∫ (g+ ĥ

)dµ =

∫gdµ +

∫ĥdµ

and so, ∫ĝdµ −

∫ĥdµ =

∫gdµ −

∫hdµ

Definition 32.10.3 Let f ∈ L1 (Ω,µ). Define∫

f dµ ≡∫

f+dµ −∫

f−dµ.

Proposition 32.10.4 The definition of∫

f dµ is well defined and if a,b are real num-bers ∫

(a f +bg)dµ = a∫

f dµ +b∫

gdµ

Proof: First of all, it is well defined because f+, f− are both no larger than | f |. There-fore,

∫f+dµ,

∫f−dµ are both real numbers. Next, why is the integral linear. First consider

the sum. ∫( f +g)dµ ≡

∫( f +g)+ dµ −

∫( f +g)− dµ

Now ( f +g)+− ( f +g)− = f +g = f+− f−+g+−g−. By Lemma 32.10.2 and Theorem32.9.1 ∫

( f +g)dµ ≡∫

( f +g)+ dµ −∫

( f +g)− dµ

=∫

( f++g+)dµ −∫

( f−+g−)dµ

=∫

f+dµ −∫

f−dµ +∫

g+dµ −∫

g−dµ

≡∫

f dµ +∫

gdµ